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Svetllana [295]
2 years ago
9

Find the surface area of this net.

Mathematics
1 answer:
Elden [556K]2 years ago
3 0

Answer:

<h2>231cm²</h2>

Step-by-step explanation:

First, let's find the surface area of both the triangles

5x3=15

So, the surface area of the triangles is 15 sq.cm

Now, let's find the surface area of the base (large rectangle in the middle)

12x8=?

10x8=80

2x8=16

80+16=96

12x8=96

So, the surface area of the base, is 96sq.cm

Now, let's find the surface area of both of the side rectangles

12x5=60

60x2=120

So, the surface area of the two side rectangles is 120sq.cm

Now, let's find the total surface area by adding all of our answers.

120+96=216

216+15=231

<h2>So hence, the surface area of this net is 231cm²</h2>
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9.3.2 Listed below are body temperatures from five different subjects measured at 8 AM and again at 12 AM. Find the values of d
kotegsom [21]

Answer:

\frac{}{d} = −0.26

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Step-by-step explanation:

Given:

Sample1:  98.1  98.8  97.3  97.5  97.9

Sample2: 98.7  99.4  97.7  97.1  98.0

Sample 1           Sample 2              Difference d

98.1                        98.7                       -0.6  

98.8                       99.4                       -0.6

97.3                        97.7                       -0.4

97.5                        97.1                         0.4

97.9                        98.0                       -0.1

To find:

Find the values of \frac{}{d} and s_{d}

d overbar ( \frac{}{d})  is the sample mean of the differences which is calculated by dividing the sum of all the values of difference d with the number of values i.e. n = 5

\frac{}{d} = ∑d/n

 = (−0.6 −0.6 −0.4 +0.4 −0.1) / 5

 = −1.3 / 5

\frac{}{d} = −0.26

s Subscript d is the sample standard deviation of the difference which is calculated as following:

s_{d} = √∑(d_{i} - \frac{}{d})²/ n-1

s_{d} =

√ (-0.6 - (-0.26))^{2} + (-0.6 - (-0.26))^{2} + (-0.4 - (-0.26))^{2} + (0.4-(-0.26))^{2} + (-0.1 - (-0.26))^{2} / 5-1

    =  √ (−0.6 − (−0.26 ))² + (−0.6 − (−0.26))² + (−0.4 − (−0.26))² + (0.4 −  

                                                                     (−0.26))² + (−0.1 − (−0.26))² / 5−1

=  \sqrt{\frac{0.1156 +  0.1156 + 0.0196 + 0.4356 + 0.0256}{4}  }

= \sqrt{\frac{0.712}{4} }

= \sqrt{0.178}

= 0.4219

s_{d} = 0.4219

Subscript d ​represent

μ_{d} represents the mean of differences in body temperatures measured at 8 AM and at 12 AM of population.

3 0
3 years ago
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